How to Solve Systems of Equations: Graphing, Substitution and Elimination

What a solution to a system means

A system of linear equations is a set of two or more linear equations that share the same unknowns. A solution is an assignment of values, such as $x = 2$ and $y = 3$, that makes every equation true simultaneously. Checking a single equation is not enough.

With two unknowns, each equation is a straight line in the plane, and a solution is a point that lies on both lines. That picture explains the three possible outcomes:

• The lines cross once: exactly one solution.
• The lines are parallel: no solution. The system is called inconsistent.
• The lines are the same line: infinitely many solutions. The equations are dependent.

Method 1: graphing

Rewrite each equation in slope-intercept form $y = mx + b$, draw both lines, and read off the intersection. Graphing is the best way to see what a system means, and it immediately shows parallel or identical lines. It is not precise, though: an intersection at $x = \frac{7}{3}$ is hard to read from a sketch. Use it to estimate, then confirm algebraically. You can draw both lines at once in the graphing calculator.

Method 2: substitution

Solve one equation for one variable, substitute that expression into the other equation, and solve the resulting one-variable equation. Substitution is the natural choice when a variable already has a coefficient of 1 or $-1$.

Always put brackets around the expression you substitute. Writing $3x + 2x - 1$ happens to work here, but substituting $y = 2x - 1$ into $3x - y = 9$ requires $3x - (2x - 1)$, and the bracket changes the sign of the $-1$.

Method 3: elimination

Add or subtract multiples of the equations so that one variable cancels. Elimination is usually fastest when both equations are in standard form $ax + by = c$.

When no coefficients match, multiply one or both equations first so that they do.

Spotting no solution or infinitely many

If both variables cancel during elimination, look at what remains.

• $x + y = 4$ and $2x + 2y = 10$. Doubling the first gives $2x + 2y = 8$; subtracting leaves $0 = 2$, which is false. The lines are parallel: no solution.
• $x + y = 4$ and $3x + 3y = 12$. Tripling the first gives exactly the second; subtracting leaves $0 = 0$. Every point on the line $x + y = 4$ is a solution.

Before you start, you can predict this from the coefficients: if $\frac{a_1}{a_2} = \frac{b_1}{b_2}$, the lines have equal slopes and the system has either no solution or infinitely many.

Cramer's rule with determinants

For a two-by-two system, determinants give each unknown directly:

For $2x + 3y = 12$ and $5x - 3y = 9$: $D = 2(-3) - 3(5) = -21$, $D_x = 12(-3) - 3(9) = -63$ and $D_y = 2(9) - 12(5) = -42$. So $x = 3$ and $y = 2$, matching elimination. If $D = 0$ the rule does not apply, which is exactly the parallel or identical case. The guide to determinants explains how to compute them for larger systems.

Three equations in three unknowns

The strategy is the same: eliminate one variable from two different pairs of equations to get a two-by-two system, solve it, then back-substitute.

Larger systems are usually solved with matrices and row reduction; the matrix calculator shows the same process.

A word problem

Two unknowns need two independent facts. If a problem gives you only one, it cannot have a unique answer.

Choosing a method and avoiding errors

• Use substitution when a variable is already isolated or has coefficient 1.
• Use elimination when the equations are in standard form, especially if coefficients already match or are opposites.
• Use Cramer's rule for a quick two-by-two answer, or when you need a formula for each unknown.
• Always check the final pair in both original equations, since an arithmetic slip can produce a point that fits only one of them.

Compare your working with the systems of equations calculator, which shows each elimination or substitution step. If you need a refresher on the single-equation steps used inside every method, see how to solve linear equations.

Which method should you use?

All the methods give the same solution, so choose the one that keeps the arithmetic simplest. Substitution is best when one equation already has a variable on its own, such as $y = 2x - 1$, or when a variable has a coefficient of 1 or $-1$. Elimination is usually faster when both equations are in standard form $ax + by = c$ and no variable is isolated, because it avoids fractions until the very end.

Graphing is useful for seeing the situation: two lines that cross, two parallel lines, or the same line drawn twice. It is less reliable for exact answers unless the intersection has integer coordinates. Cramer's rule and matrix methods suit systems with three or more variables, and they are what calculators and computer software use internally.

For nonlinear systems, such as a line and a parabola, substitution is almost always the right choice. Solving $y = x^2$ and $y = x + 2$ gives $x^2 = x + 2$, so $x^2 - x - 2 = 0$ and $(x - 2)(x + 1) = 0$. The intersection points are $(2, 4)$ and $(-1, 1)$, and both satisfy each equation.

Checking a solution

A solution of a system must satisfy every equation, not just the one you used last. Substitute your values into each original equation. For the second elimination example, $(2, 5)$ gives $3(2) + 2(5) = 16$ and $4(2) - 5(5) = -17$, so both equations hold. A value that satisfies one equation but not the other points to an arithmetic slip during elimination.

Write the check on your paper. It costs less than a minute and examiners frequently award credit for a verified answer.

The systems of equations calculator solves two and three variable systems with each step shown, which is a convenient way to compare your working.

Frequently asked questions